The rate of rotation of a solid disk with a radius of #2 m# and mass of #2 kg# constantly changes from #5 Hz# to #9 Hz#. If the change in rotational frequency occurs over #4 s#, what torque was applied to the disk?

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Answer 1 · 2018-01-21T21:13:17

#8pi mN ~~25 mN#

The applied torque is
#vectau=vecRxxvecF#
The reaction to this torque is
#vectau = I vec alpha#

The magnitude of the torque is:

#tau = I alpha#

where #I=1/2MR^2 and alpha= (omega_2- omega_1)/(Deltat)#

#because omega=2pi f#

#tau = I alpha= 1/cancel2MR^2*(cancel2pi(f_2-f_1))/(Deltat) #

Substitute #M=2 kg, R= 2m, f_2=9 Hz, f_1= 5Hz; Deltat =4s#

#tau = 2kg*(2m)^2*pi*((9-5)Hz)/(4s)= 8pi m*(kgm)/s^2~~25mN#

To generate this toque, you need a force #tau/R=(25mN)/(2m)#= 12.5N